A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the bait and hook reach a maximum height of $2.9 \mathrm {~m}$ What was the initial velocity he launched the bait with? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible.
A) 18 m/s
B) 7.9 m/s
C) 7.6 m/s
D) 6.3 m/s

Question

Quizplus · Accepted Answer

Explanation:The vertical component of the initial velocity is:$$v_{y_i} = v_i \sin 25^\circ$$The maximum height reached by the bait is:$$h = \frac{v_{y_i}^2}{2g}$$Substituting the given values, we get:$$2.9 \mathrm{~m} = \frac{(v_i \sin 25^\circ)^2}{2(9.8 \mathrm{~m/s^2})}$$Solving for $v_i$, we get:$$v_i = \frac{\sqrt{2.9 \mathrm{~m} \times 2 \times 9.8 \mathrm{~m/s^2}}}{\sin 25^\circ} = 18 \mathrm{~m/s}$$

A Fisherman Casts His Bait Toward the River at an Angle